Logic colloquium '80. Papers intended for the European by D.van Dalen, etc.

By D.van Dalen, etc.

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So (A-P) can be rewritten as consisting 0+-+3p&Q0 If 0 has free variables x, then 0(x) is a predicate on some preset X (over which x is supposed to range); that means that 0 determines a function which assigns to each x a preset Q0(x) of proofs of 0(x), so (A-P) can be written 0(x)~P&Q0(x). In contrast to the problem of unbounded quantification, nobody seems to object to the idea that there are certain operations which can apply to any objects whatFor example, given any two objects x and y, we can form a pair (x,y); the ever.

J. BEESON 30 given by finite descriptions. This is certainly the case with intuitionism, where choice sequences by definition have no finite descriptions, and is also a view consistent with Bishop's phi losophy. In this case, it is not clear how to construct or define an equality relation on the entire univers (of constructed objects). See [27] for a long discussion. (ii i) Even if one admits that every mathematical object can be described in natural language, because of the inherent vagueness of such descriptions and because of their inherent context-dependency, Greenleaf maintains that the construction of (i) is not well-defined.

Like the recursion theorem, it has a very short proof which is difficult to remember: Theorem (Diaconescu [19]): The axiom of choice (in the fo~: every set of nonempty sets has a choice function choosing an element of each of its members) implies the law of the excluded middle, using separation and extensionality, Proof: Let 0 be a given statement; we want to derive 0 v 10. Define A and B by A~ {ntN: n~Ov(n~1 & 0)} B~ {ntN: n~lv(n~O & 0)} Then A and B are non-empty sets. f(A)tA and f(B)tB. or f(A)#f(B).

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